Number of unknowns - Coordinate Transforms

thehangedman
Messages
68
Reaction score
2
In general relativity, what are the total number of unknowns for a generic coordinate transform? Is it just 4 * 4 = 16? Is there a way to break those down into combinations of types, such as boosts, rotations, reflections (parity?), etc, or is it just left wide open from an interpretive standpoint? My feeling is the answer is in fact 16 unknown functions of space-time and that the actual interpretation can't really be broken out like we do in SR (Lorentz)...

Your help is greatly appreciated...
 
Physics news on Phys.org
It is an infinite number of unknowns. Even if you only had a 1D manifold, there are an infinite number of degrees of freedom.
 
DaleSpam said:
It is an infinite number of unknowns. Even if you only had a 1D manifold, there are an infinite number of degrees of freedom.

I made a mistake in my question. I know that functions have an unlimited number of degrees of freedom, I meant to ask how many functions are involved in a generic coordinate transform in R^4. My guess is 16, since there are two indexes in the transformation matrix, each running over 4 values, but wanted clarity since there might be symmetries that eliminate elements (though I'm guessing not).
 
I think it is only 4 unknown functions. You can always write it as:
x'_0=f_0(x_0,x_1,x_2,x_3)
x'_1=f_1(x_0,x_1,x_2,x_3)
x'_2=f_2(x_0,x_1,x_2,x_3)
x'_3=f_3(x_0,x_1,x_2,x_3)
 
May be you are talking about gμv which seems to have 16 components.But it is symmetric,which will eliminate 6 so there will be only 10 independent components.
 
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Abstract The gravitational-wave signal GW250114 was observed by the two LIGO detectors with a network matched-filter signal-to-noise ratio of 80. The signal was emitted by the coalescence of two black holes with near-equal masses ## m_1=33.6_{-0.8}^{+1.2} M_{⊙} ## and ## m_2=32.2_{-1. 3}^{+0.8} M_{⊙}##, and small spins ##\chi_{1,2}\leq 0.26 ## (90% credibility) and negligible eccentricity ##e⁢\leq 0.03.## Postmerger data excluding the peak region are consistent with the dominant quadrupolar...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
Back
Top